Problem: Divide the following complex numbers. $\dfrac{-8-6i}{3+i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${3-i}$. $ \dfrac{-8-6i}{3+i} = \dfrac{-8-6i}{3+i} \cdot \dfrac{{3-i}}{{3-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(-8-6i) \cdot (3-i)} {3^2 - (i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(-8-6i) \cdot (3-i)} {(3)^2 - (i)^2} $ $ = \dfrac{(-8-6i) \cdot (3-i)} {9 + 1} $ $ = \dfrac{(-8-6i) \cdot (3-i)} {10} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-8-6i}) \cdot ({3-i})} {10} $ $ = \dfrac{{-8} \cdot {3} + {-6} \cdot {3 i} + {-8} \cdot {-1 i} + {-6} \cdot {-1 i^2}} {10} $ $ = \dfrac{-24 - 18i + 8i + 6 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{-24 - 18i + 8i - 6} {10} = \dfrac{-30 - 10i} {10} = -3-i $